An object of mass 2 kg attached to a spring is moved to a distance x = 10 m from its equilibrium position. The spring constant k = 1 N `m^(-1)` and assume that the surface is frictionless.
What is the force that acts on the object when the mass crosses the equilibrium position and extremum position `x=pm10m`.

Explanation

Since the restoring spring force is F = -kx, when the object crosses the equilibrium position, it experiences no force. Note that at equilibrium position, the object moves very fast. When the object is at x = `pm10` m (elongation) the force F = - kx
F = - (1) (10) = - 10 N. Here the negative sign implies that the force is towards equilibrium i.e., towards negative x-axis and when the object is at x = - 10m (compression), it experiences a forces F = - (1) (-10) = `pm10` N. Here the positive sign implies that the force points towards positive x-axis.
`" "` The object comes to momentary rest at `x=pm10m` even through it experiences a maximum force at both these points.